The statement you want to prove is for all $n\in\mathbb{N}$ it holds that $n^2\leq n!$ (you called this $P(n)$. So lets first prove $P(4)$ i.e. $4^2\leq 4!$ but since $16\leq 24$ this is clear. So lets assume $P(n)$ and prove $P(n+1)$.

First note that for $n\geq 2$ it holds that
$$ 0\leq (n-1)^2+(n-2)=n^2-2n+1+n-2=n^2-n-1 $$
which is equivalent to $n+1\leq n^2$ which gives

$$ (n+1)^2=(n+1)(n+1)\leq (n+1)n^2 $$

by induction hypothesis (i.e. $P(n)$) the term $n^2$ in the last expression is smaller or equal $n!$ so we can continue:
$$ (n+1)n^2\leq (n+1)n! = (n+1)! $$
which is the statement we wanted to prove.

My answer is very extensive and explicit. But maybe, you now get a better understanding of what you have to do in general, when you want to prove something by induction.

One way to prove this is to strengthen your induction hypothesis slightly: assume $k^2+k\leq k!$ instead of $k^2\leq k!$. You have a base case for this starting with $k=4$, where $k^2+k=20$ and $k!=24$.

Then we want to show $(k+1)^2+k+1<(k+1)!$. Divide both sides by $k+1$ to see this holds just if $k+2<k!$. But we supposed $k^2+k<k!$ and $k\geq 4$, so this holds. Then we've shown for all $n$ that $n^2+n\leq n!$, which implies the conclusion.